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comment On the derivative of a Heaviside step function being proportional to the Dirac delta function
@Joebevo you have to take into account that the area "under" the delta function is also normalized. So you always have $\int_{-\infty}^\infty \delta(x) = 1$. Therefore $\delta(x)$ and $\alpha\delta(x)$ are not the same function, because if you integrate the latter you get $\alpha$.
Nov
15
answered On the derivative of a Heaviside step function being proportional to the Dirac delta function
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accepted What is this generalized multivariable hypergeometric function?
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answered What is this generalized multivariable hypergeometric function?
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Feb
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comment What is this generalized multivariable hypergeometric function?
I have the seventh edition of this book, and the page number 1057 doesn't make sense (it's some stuff about integration). In chapter 9 they mention some multivariable hypergeometric functions, but these are either the Lauricella ones or functions of only two variables.. Do you perhaps have a newer edition?
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asked What is this generalized multivariable hypergeometric function?